Suppose that $A=\left(a_{ij}\right)\in M_n(\mathbb{R})$ is a symmetric matrix with positive entries. Let $$\lambda_1\geq\lambda_2 \dots \geq \lambda_n$$ be the eigenvalues of $A$. Thus there exists an orthogonal matrix $C$ such that \begin{equation*} A= C^t\begin{pmatrix} \lambda_{1} & 0 & \cdots & 0 \\ 0 & \lambda_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_{n} \end{pmatrix}C . \end{equation*}
My question is when I change some $\lambda_i$'s slightly, can we still obtain a symmetric matrix with positive entries under the same transformation continuous.
For example, we consider changing one eigenvalue $\lambda_1$ to $\lambda_1+\epsilon$. Denote \begin{equation*} B=\left(b_{ij}\right):= C^t\begin{pmatrix} \lambda_{1}+\epsilon & 0 & \cdots & 0 \\ 0 & \lambda_{2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_{n} \end{pmatrix}C. \end{equation*} Can we find a sufficiently small $\epsilon$ such that $B$ is still a symmetric matrix with positive entries. I also wonder what the distance between $A$ and $B$, that is, can $$\sum_{i,j}(a_{ij}-b_{ij})^2$$ tends to $0$ as $\epsilon$ tends to $0$.
I don't know whether there is a general solution to this question. I think this can be right but I don't know how to prove it. Any help will be appreciated.:)